Chapter 1
Limit Cycle and Its Perturbations
1.1 Basic notations and facts
Consider a planar system deˉned on a region G?R2 of the form
x_=f(x);(1:1:1)
where f:G ?! R2 is a Cr function,r > 1.Then for any point x02 G (1.1.1) has a
unique solution '(t;x0) satisfying '(0;x0)=x0:Let't(x0)='(t;x0):The family
of the transformations 't:G ?! R2 satisˉes the followingproperties
(i) '0=Id;
(ii) 't+s='t ± 's:
The function ' is called the °ow generated by (1.1.1) or by thevector ˉeld f.Let
I(x0) denote the maximal interval of deˉnition of '(t;x0) int.If x0 2 G is such
that '(t;x0) is constant for all t 2 I(x0),then f(x0)=0.In thiscase,x0 is called
a singular point of (1.1.1).A point that is not singular iscalled a regular point.
For any regular point x0 2 G,the solution '(t;x0) determines twoplanar curves
as follows
°+(x0)=f'(t;x0):t 2 I(x0);t > 0g;°?(x0)=f'(t;x0):t 2 I(x0);t6 0g;
which are called the positive,negative orbit of (1.1.1) throughx0 respectively.The
union °(x0)=°+(x0)[°?(x0) is called the orbit of (1.1.1) throughx0.The theorem
about the existence and uniqueness of solutions ensures thatthere is one and only
one orbit through any point in G.Thus,it is easy to prove thatany di?erent orbits
do not intersect each other.A periodic orbit of (1.1.1) is anorbit that is a closed
curve.The minimal positive number satisfying '(T;x0)=x0 is saidto be the period
of the periodic orbit °(x0).Obviously,°(x0) is a periodic orbitof period T if and
only if the corresponding representation '(t;x0) is a periodicsolution of the same
period.
Deˉnition 1.1.1 A periodic orbit of (1.1.1) is called a limitcycle if it is the
only periodic orbit in a neighborhood of it.In other words,alimit cycle is an isolated
periodic orbit in the set of all periodic orbits.
Now let us assume that (1.1.1) has a limit cycle L:x=u(t);0 6 t6 T.
Since (1.1.1) is autonomous,for any given point p 2 L we maysuppose p=u(0),
and hence,u(t)='(t;p).Further,for deˉniteness,let L be orientedclockwise.
Introduce a unit vector below
Z0 =
1
jf(p)j
(?f2(p);f1(p))T:
Then there exists a cross section l of (1.1.1) which passesthrough p and is parallel
to Z0.Clearly,a point x0 2 l near p can be written as x0=p + aZ0with a =
(x0?p)TZ0 2 R small.
Lemma 1.1.1 There exist a constant " > 0 and Cr functions Pand ?:
(?";") ?! R with P(0)=0 and?(0)=T such that
'(? (a);p + aZ0)=p + P(a)Z0 2 l;jaj < ":(1:1:2)
Proof Deˉne Q(t;a)=[f(p)]T('(t;p + aZ0)?p).We have
Q(T;0)=0;Qt(T;0)=jf(p)j2 > 0:
Note that Q is Cr for (t;a) near (T;0):The implicit functiontheorem implies that
a Cr function?(a)=T + O(a) exists satisfying
Q(? (a);a)=0 or [f(p)]T('(? (a);p + aZ0)?p)=0:
It follows that the vector '(? (a);p + aZ0)?p is parallel toZ0.Hence,it can be
rewritten as '(? (a);p + aZ0)?p=P(a)Z0,where
P(a)=ZT
0 ('(? (a);p + aZ0)?p):(1:1:3)
It is obvious that P 2 Cr for jaj small with P(0)=0.This endsthe proof.
The above proof tells us that the function?is the time of theˉrst return to l.
By Deˉnition 1.1.1,the periodic orbit L is a limit cycle if andonly if P(a) 6= a for
jaj > 0 su±ciently small.
Deˉnition 1.1.2 The function P:(?";") ! R deˉned by (1:1:2) iscalled a
Poincar?e map or return map of (1.1.1) at p 2 l.
For convenience,we sometimes use the notation P:l ?! l.
Deˉnition 1.1.3 The limit cycle L is said to be outer stable(outer unstable)
if for a > 0 su±ciently small,
a(P(a)?a) < 0(> 0):
The limit cycle L is said to be inner stable (inner unstable) ifthe inequality above
holds for ?a > 0 su±ciently small.A limit cycle is calledstable if it is both inner
and outer stable.A limit cycle is called unstable if it is notstable.
For example,if L is stable,then the orbits near it behave likethe phase portrait
as shown in Figure 1.1.1.
Let Pk(a) denote the kth iterate of a under P.It is evident thatfPk(a)g is
monotonic in k and Pk(a) > 0(< 0) for a > 0(<0).Thus,it is easy to see that L
is outer stable if and only if Pk(a) ! 0 as k ! 1 for all a >0 su±ciently small.
Similar conclusions hold for outer unstable,inner stable andinner unstable cases.
Remark 1.1.1 If the limit cycle L is oriented anti-clockwise wecan deˉne its
stability in a similar manner by using the Poincar?e map Pdeˉned by (1.1.2).For
instance,it is said to be inner stable (inner unstable) ifa(P(a)?a) < 0(> 0) for
a > 0 su±ciently small.
Deˉnition 1.1.4 The limit cycle L is said to be hyperbolic or ofmultiplicity
one if P0(0) 6= 1.It is said to have multiplicity k,2 6 k 6 r;ifP0(0)=1;P(j)(0) =
0;j=2;￠ ￠ ￠ ;k?1;P(k)(0) 6= 0:
By Deˉnition 1.1.3,one can see that L is stable (unstable) ifjP0(0)j < 1 (> 1):
Example 1.1.1 Consider a system given by
_ x1=x1?x2?x1(x2
1 + x2
2);
_ x2=x1 + x2?x2(x2
1 + x2
2):
(1:1:4)
The system has the form
r_=r(1?r2);μ_=1
in polar coordinates (r;μ) with x=(r cos μ;r sin μ).Thus,one canˉnd (1.1.4) has
a °ow of the form
'(t;x0)=r(t)(cos μ(t);sin μ(t))T;(1:1:5)
where
r(t)=r0(r2
0 + (1?r2
0)e?2t)?1
2 ;μ(t)=t + μ0;
x0=r0(cos μ0;sin μ0)T;r0 > 0;0 6 μ0 < 2:
For p=(1;0)T,we have a periodic orbit L=f(x1;x2)Tjx2
1 + x2
2=1g which has a
representation
L:x='(t;p)=(cos t;sin t)T;0 6 t 6 2;
with Z0=(?1;0)T.Then p+aZ0=(1 ?a;0)T.Hence,x0=p+aZ0 if and onlyif
r0=1?a,μ0=0.
Taking l=f(x1;0)Tjx1 > 0g.Then noting that?(0)=2,by (1.1.5)we have
'(? (a);p + aZ0) 2 l if and only if?(a)=2 for a <1.Therefore,
'(? (a);p + aZ0)=(1?a)[(1?a)2 + (2a?a2)e?4]?1
2 (1;0)T:
It follows from (1.1.3) that
P(a)=1?(1?a)[(1?a)2 + (2a?a2)e?4]?1
2
= ae?4 + O(a2)
for jaj small.By Deˉnition 1.1.3,the limit cycle L isstable.
Next,we give formulas for P0(0) and P00(0).For thepurpose,let
v(μ)=u0(μ)
ju0(μ)j
= (v1(μ);v2(μ))T;Z(μ)=(?v2(μ);v1(μ))T;
and introduce a transformation of coordinates of the form
x=u(μ) + Z(μ)b;0 6 μ 6 T;jbj < ":(1:1:6)
Lemma 1.1.2 The transformation (1:1:6) carries (1:1:1) into thesystem
dμ
dt
= 1 + g1(μ;b);
db
dt
= A(μ)b + g2(μ;b);(1:1:7)
where
A(μ)=ZT(μ)fx(u(μ))Z(μ)=trfx(u(μ)) ?
d
dμ
ln jf(u(μ))j;
g1(μ;b)=h(μ;b)[f(u(μ) + Z(μ)b)?f(u(μ))]?h(μ;b)Z0(μ)b;
g2(μ;b)=ZT(μ)[f(u(μ) + Z(μ)b)?f(u(μ))?fx(u(μ))Z(μ)b];
h(μ;b)=(jf(u(μ))j + vT(μ)Z0(μ)b)?1vT(μ);
and trfx(u(μ)) denotes the trace of the matrix fx(u(μ)),which iscalled the divergence
of the vector ˉeld f evaluated at u(μ):
Proof By (1.1.6) and (1.1.1) we have
(u0 + Z0b)
d μ
d t
+ Z
d b
d t
= f(u + Z b):(1:1:8)
In order to obtain (1.1.7) we need to solve
d μ
d t
and
d b
d t
from (1.1.8).First,multiplying
(1.1.8) by vT from the left-hand side and using
vTZ=0;vTf(u)=vTu0=ju0j=jf(u)j;
we can obtain
dμ
dt
= [jf(u)j + vTZ0b]?1vTf(u + Zb)=h(μ;b)f(u + Zb):
Note that
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